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Question:
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Let the vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is a unit vector if the angle between $\vec{a}$ and $\vec{b}$ is
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{2}$.
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{2}$.
Solution:
1452 Upvotes
Verified Answer
Given $|\vec{a} \times \vec{b}|=1,|\vec{a}|=3,|\vec{b}|=\frac{\sqrt{2}}{3}$
Now $\quad|(\vec{a} \times \vec{b})=| \vec{a} \| \vec{b} \mid \sin \theta$
$$
\Rightarrow \quad \sin \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}
$$
Hence correct, option is (b).
Now $\quad|(\vec{a} \times \vec{b})=| \vec{a} \| \vec{b} \mid \sin \theta$
$$
\Rightarrow \quad \sin \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}
$$
Hence correct, option is (b).
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